Question

. A metal M of atomic weight 54.9 has a density of
7.42 g cm ^-3. Calculate the volume occupied and the radius
of the atom of this metal assuming it to be sphere.​

Answer 1

Answer:

Explanation:

a container used for transportation has capacity 34.5metercube the length and bradth of conainer are 5m and 2.3m find the height of container

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

We are given the density and mass of the metal. Using this, we can calculate the volume occupied by the metal. Also by Avogadro's Constant we can calculate the volume of each atom. And then using the formula of volume of sphere, we can find the radius of atom of metal.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{Density\;=\;\bf{\dfrac{Mass}{Volume}}}}$

$\\\;\boxed{\sf{Volume\;of\;Sphere\;=\bf{\dfrac{4}{3}\:\pi r^{3}}}}$

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★ Solution :-

Given,

» Density of metal = 7.42 g/cm³

» Mass of Metal = 54.9 g

» No. of atoms in metal = 6.022 × 10²³

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~ For the Volume Occupied by Metal ::

$\\\;\;\sf{:\rightarrow\;\;Density\;=\;\bf{\dfrac{Mass}{Volume}}}$

$\\\;\;\sf{:\rightarrow\;\;Volume\;=\;\bf{\dfrac{Mass}{Density}}}$

$\\\;\;\sf{:\rightarrow\;\;Volume\;=\;\bf{\dfrac{54.9}{7.42}}}$

$\\\;\;\sf{:\rightarrow\;\;Volume\;of\;metal\;=\;\bf{\green{7.40\;\;cm^{3}}}}$

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~ For the Volume of each Atom of metal ::

We know that,

✒ Number of atoms = 6.022 × 10²³

Then,

$\\\;\;\sf{:\Longrightarrow\;\;Volume\;of\;atom\;=\;\bf{\dfrac{Volume\;of\;Metal}{No.\;of\;Atoms}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Volume\;of\;atom\;=\;\bf{\dfrac{7.40}{6.022\;\times\;10^{23}}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Volume\;of\;atom\;=\;\bf{\pink{1.23\;\times\;10^{-23}\;\;cm^{3}}}}$

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~ For Radius of each Spherical Atom ::

• Let the radius of the atom be 'r' cm.

$\\\;\;\sf{:\Longrightarrow\;\;Volume\;of\;Sphere\;=\bf{\dfrac{4}{3}\:\pi r^{3}}}$

$\\\;\;\sf{:\Longrightarrow\;\;Volume\;of\;atom\;=\bf{\dfrac{4}{3}\:\pi r^{3}}}$

$\\\;\;\sf{:\Longrightarrow\;\;\dfrac{4}{3}\:\times\:\dfrac{22}{7}\:r^{3}\;=\;\bf{1.23\;\times\;10^{-23}}}$

$\\\;\;\sf{:\Longrightarrow\;\;r^{3}\;=\;\bf{\dfrac{7\;\times\;3\;\times\;1.23\;\times\;10^{-23}}{4\;\times\;22}}}$

$\\\;\;\sf{:\Longrightarrow\;\;r^{3}\;=\;\bf{0.3\;\times\;10^{-23}}}$

$\\\;\;\sf{:\Longrightarrow\;\;r\;=\;\bf{\sqrt[3]{0.3\;\times\;10^{-23}}}}$

$\\\;\;\sf{:\Longrightarrow\;\;r\;=\;\bf{0.67\;\times\;2.2\;\times\;10^{-8}}}$

$\\\;\;\sf{:\Longrightarrow\;\;r\;=\;\bf{\red{1.474\;\times\;10^{-8}\;cm}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;radius\;\;of\;\;atom\;\;=\;\bf{\purple{1.474\;\times\;10^{-8}\;cm}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Molarity\;=\;\dfrac{No.\;of\;Moles}{Volume\;of\;Solution}}$

$\\\;\sf{\leadsto\;\;Molarity\;=\;\dfrac{No.\;of\;Moles}{Mass\;of\;Solvent}}$

$\\\;\sf{\leadsto\;\;Mass\;per\;cent\;=\;\dfrac{Mass\;of\;Solute}{Mass\;of\;Solution}\;\times\;100}$